EM 1110-2-1100 (Part II)
30 Apr 02
EXAMPLE PROBLEM II-6-4
Find:
Determine the wave height in an inlet channel that is 5.0 m (16.4 ft) deep.
Given:
A 5.0-sec, 1.0-m (3.28-ft) wave is entering an inlet having a 1.0 m/s (3.3 ft/s) ebb current. The angle
between current and wave orthogonal is 180 deg.
Solution:
Ω = (2π/T) (dT/g)1/2
F = (V cos θ)/(gT)1/2
T = wave period = 5 s
dT = time-averaged water depth = 5.0 m
g = 9.8 m/s2
V = horizontal current speed = 1.0 m/s
θ = angle between horizontal velocity vector and horizontal wave vector = 180o
Therefore:
1/2
2 (3.14)
5
Ω'
' 0.90
5.0
9.8
1.0 (cos 180)
F'
' &0.14
1/2
( 9.8 @ (5) )
From Figure II-6-34, RH = 1.35
Wave height HA, originally 1 m, is modified by the current to a wave height of:
H = (RH) HA = 1.35 (1.0) = 1.35 m
II-6-42
Hydrodynamics of Tidal Inlets
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